Expert: Sherry Wallin - 2/16/2011

Question
QUESTION: 6 tan^2 x – 2 cos^2 x  =  cos2x  if  cos2x  = ?

ANSWER: Vipul~

When x = pi/6 cos 2x = pi/3 and cos pi/3 = 1/2
6*tan^2(pi/6)-2*cos^2(pi/6) = 6*((sqrt3)/3)^2-2*((sqrt3)/2)^2 = 6*(3/9)-2*(3/4) = 2-(3/2) = 1/2

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Dear Sir,
Will you please provide a mathematical solution of above problem for cos2x in place of assuming x = pi/6.With regards and in anticipation,
truly yours
Vipul.

Answer
Vipul~

Start by moving cos 2x to the left getting 6 tan^2 x – 2 cos^2 x  -  cos2x = 0

Now change the tan^2x to sin^2x/(1-sin^2x) and cos^2x to 1-sin^2x and cos 2x to 1-2sin^2x:

6 sin^2x/(1-sin^2x) -2(1-sin^2x) -(1-2sin^2x) = 0

Now multiply both sides by (1-sin^2x) getting:

6 sin^2x -2(1-sin^2x)^2 -(1-sin^2x)(1-2sin^2x) = 0

It is easier for you to see what is going on if we make a simple substitution of letting u = sin^2:

6u-2(1-u)^2-(1-u)(1-2u) = 0

6u-2(1-2u+u^2)-(1-3u+2u^2) = 0

6u - 2 + 4u - 2u^2 - 1 + 3u - 2u^2 = 0

-4u^2 + 13u - 3 = 0

4u^2 - 13u + 3 = 0

4u^2 - u - 12u + 3 = 0
u(4u - 1)-3(4u - 1) = 0
u - 3 = 0 or 4u - 1 = 0
u = 3 or u = 1/4

Now put back in what u is:
sin^2x = 3 => sin x = sqrt3 > 1 => No solution since |sin x| <= 1
or
sin^2x = 1/4 => sin x = 1/2 => x = sin^-1 (1/2) => x = pi/6

Math Prof